Question

integrate

Answer 1

which function

Answer 2

\[\Large \int\limits_{0}^{\infty}\frac{\sin^{4}(x)}{x^4}\]

Answer 3

Hint \[\Large \int\limits_{0}^{\infty} \frac{\sin^{2}(x)}{x^2}=\frac{\pi}{2}\]

Answer 4

@phi

Answer 5

residue theorem gives it (2*pi)/3

Answer 6

\[res[f(z)]=(-2 \pi i)\lim_{z \rightarrow 0}\frac{ 1 }{ [4-1]! }\frac{ d^3 }{ dz^3 }(z)^4\]

Answer 7

with some negetive exponentials of x

Answer 8

we can consider here two contours one for positive another for negetive
for +ve contour lies above real plane

Answer 9

and there is a pole at z=0 since the function becomes infinity at z=0

Answer 10

so there is'nt any residue for positive contour which lies in the real plane
so \[res[pos. exponentials]=0\]
and so we just need to figure out residue for negetive exponentials which lies below the real plane i think!

Answer 11

@sidsiddhartha I was wondering if there's a way to use the formula
\[PV\int_{-\infty}^\infty \frac{P(x)}{Q(x)}\sin x~dx=\\
~~~~~~~\text{Im}\left\{2\pi i\left(\sum\text{Res}_{\text{Im}z\ge0}\frac{P(z)}{Q(z)}e^{iz}+\frac{1}{2}\sum\text{Res}_{\text{Im}z=0}\frac{P(z)}{Q(z)}e^{iz}\right)\right\}\]

Answer 12

Which of the following best describes the graph of f(x) = x^2 - 7x + 10?

Minimum at (1.5, -12.25) with intercepts at (5, 0) and (-2, 0)

Minimum at (-1.5, -12.25) with intercepts at (-5, 0) and (2, 0)

Minimum at (-3.5, -2.25) with intercepts at (-5, 0) and (-2, 0)

Minimum at (3.5, -2.25) with intercepts at (5, 0) and (2, 0)

Answer 13

hey @SithsAndGiggles i'm not sure how to use this :(

Answer 14

\[\begin{array} \\


I(a) & = \int_0^{\infty } \dfrac{\sin ^4(ax)}{x^4} dx \\
I'(a) & = 4\int_0^{\infty } \dfrac{\sin ^3(ax) \cos(ax)}{x^3} dx \\
I''(a) & = 4\int_0^{\infty } \dfrac{3\sin ^2(ax) \cos^2(ax) - \sin^4(ax)}{x^2} dx \\

I''(a) & = 4\int_0^{\infty } \dfrac{\frac{3}{4}\sin ^2(2ax) - \sin^2(ax) + \frac{1}{4}\sin^2(2ax)}{x^2} dx \\

I''(a) & = 4\int_0^{\infty } \dfrac{\sin ^2(2ax) - \sin^2(ax) }{x^2} dx \\
I''(a) & = 4(2a)^2 \int_0^{\infty } \dfrac{\sin ^2(2ax) }{(2ax)^2} dx - 4a^2\int_0^{\infty } \dfrac{\sin ^2(ax) }{(ax)^2} dx \\

I''(a) & = 4(2a)^2\frac{\pi}{2}\times \frac{1}{2a}- 4a^2\frac{\pi}{2}\times \frac{1}{a} \\


\end{array} \]

Answer 15

I was thinking a power reduction formula might work, but I too don't know what to do with that:
\[\sin^4x=\frac{3}{8}-\frac{1}{2}\cos2x+\frac{1}{8}\cos4x\]
Basically, the formula says the principal value integral is \(2\pi i\) times the sum of the residues in the upper half plane and half the sum of the residues on the real line. Your residue computation does just that - there are no complex singularity points, so you just have
\[\pi i~\text{Res}_{z=0}f(z)\]

Answer 16

yes that will make it easier now we just have to convert them into exponential is'nt it?